tank volume required for a target detention time in a storage tank Given flow rate (Q) = 16,350 m^3/day; hydraulic detention time (theta) = 7.3000 day, determine the tank volume (V) in m^3.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• Everything except V is given, so isolate V symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for V:

   $$V = Q\,\theta$$

3. Step 3 — List the givens: flow rate (Q) = 16,350 m^3/day, hydraulic detention time (theta) = 7.3000 day.

4. Step 4 — Substitute the given values:

   $$V = 16350\,7.3000$$

5. Step 5 — Evaluate:

   V = 119355\ \text{m^3}

6. Step 6 — Check: returning V = 119,355 m^3 to

   $$V = Q \times \theta$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 238,710 — kept a factor of two that cancels in the correct rearrangement.
• 59,678 — dropped that same factor in the other direction.
• 131,291 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume

tank volume sizing for an equalization basin Given hydraulic detention time (theta) = 5.1000 day; tank volume (V) = 149,316 m^3, determine the flow rate (Q) in m^3/day.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• Everything except Q is given, so isolate Q symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for Q:

   $$Q = \dfrac{V}{\theta}$$

3. Step 3 — List the givens: hydraulic detention time (theta) = 5.1000 day, tank volume (V) = 149,316 m^3.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{149316}{5.1000}$$

5. Step 5 — Evaluate:

   Q = 29278\ \text{m^3/day}

6. Step 6 — Check: returning Q = 29,278 m^3/day to

   $$V = Q \times \theta$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 58,555 — kept a factor of two that cancels in the correct rearrangement.
• 14,639 — dropped that same factor in the other direction.
• 32,205 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume

detention tank volume calculated from plant flow rate Given flow rate (Q) = 2,220 m^3/day; tank volume (V) = 218,303 m^3, determine the hydraulic detention time (theta) in day.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• Everything except theta is given, so isolate theta symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for theta:

   $$\theta = \dfrac{V}{Q}$$

3. Step 3

   $$List the givens: flow rate (Q) = 2,220 m^3/day, tank volume (V) = 218,303 m^3$$

4. Step 4 — Substitute the given values:

   $$\theta = \dfrac{218303}{2220}$$

5. Step 5 — Evaluate:

   $$\theta = 98.3347\ \text{day}$$

6. Step 6 — Check: returning theta = 98.3347 day to

   $$V = Q \times \theta$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 196.7 — kept a factor of two that cancels in the correct rearrangement.
• 49.1673 — dropped that same factor in the other direction.
• 108.2 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume

tank volume required for a target detention time in a storage tank Given flow rate (Q) = 5,000 m^3/day; hydraulic detention time (theta) = 19.0000 day, determine the tank volume (V) in m^3.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• Everything except V is given, so isolate V symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for V:

   $$V = Q\,\theta$$

3. Step 3 — List the givens: flow rate (Q) = 5,000 m^3/day, hydraulic detention time (theta) = 19.0000 day.

4. Step 4 — Substitute the given values:

   $$V = 5000\,19.0000$$

5. Step 5 — Evaluate:

   V = 95000\ \text{m^3}

6. Step 6 — Check: returning V = 95,000 m^3 to

   $$V = Q \times \theta$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 190,000 — kept a factor of two that cancels in the correct rearrangement.
• 47,500 — dropped that same factor in the other direction.
• 104,500 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume

tank volume sizing for an equalization basin Given hydraulic detention time (theta) = 18.0000 day; tank volume (V) = 199,020 m^3, determine the flow rate (Q) in m^3/day.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• Everything except Q is given, so isolate Q symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for Q:

   $$Q = \dfrac{V}{\theta}$$

3. Step 3 — List the givens: hydraulic detention time (theta) = 18.0000 day, tank volume (V) = 199,020 m^3.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{199020}{18.0000}$$

5. Step 5 — Evaluate:

   Q = 11057\ \text{m^3/day}

6. Step 6 — Check: returning Q = 11,057 m^3/day to

   $$V = Q \times \theta$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 22,113 — kept a factor of two that cancels in the correct rearrangement.
• 5,528 — dropped that same factor in the other direction.
• 12,162 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume

detention tank volume calculated from plant flow rate Given flow rate (Q) = 11,260 m^3/day; tank volume (V) = 21,977 m^3, determine the hydraulic detention time (theta) in day.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• Everything except theta is given, so isolate theta symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for theta:

   $$\theta = \dfrac{V}{Q}$$

3. Step 3

   $$List the givens: flow rate (Q) = 11,260 m^3/day, tank volume (V) = 21,977 m^3$$

4. Step 4 — Substitute the given values:

   $$\theta = \dfrac{21977}{11260}$$

5. Step 5 — Evaluate:

   $$\theta = 1.9518\ \text{day}$$

6. Step 6 — Check: returning theta = 1.9518 day to

   $$V = Q \times \theta$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 3.9036 — kept a factor of two that cancels in the correct rearrangement.
• 0.9759 — dropped that same factor in the other direction.
• 2.1470 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume

tank volume required for a target detention time in a storage tank Given flow rate (Q) = 15,100 m^3/day; hydraulic detention time (theta) = 9.0000 day, determine the tank volume (V) in m^3.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• Everything except V is given, so isolate V symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for V:

   $$V = Q\,\theta$$

3. Step 3 — List the givens: flow rate (Q) = 15,100 m^3/day, hydraulic detention time (theta) = 9.0000 day.

4. Step 4 — Substitute the given values:

   $$V = 15100\,9.0000$$

5. Step 5 — Evaluate:

   V = 135900\ \text{m^3}

6. Step 6 — Check: returning V = 135,900 m^3 to

   $$V = Q \times \theta$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 271,800 — kept a factor of two that cancels in the correct rearrangement.
• 67,950 — dropped that same factor in the other direction.
• 149,490 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume

tank volume sizing for an equalization basin Given hydraulic detention time (theta) = 0.9000 day; tank volume (V) = 79,559 m^3, determine the flow rate (Q) in m^3/day.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• Everything except Q is given, so isolate Q symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for Q:

   $$Q = \dfrac{V}{\theta}$$

3. Step 3 — List the givens: hydraulic detention time (theta) = 0.9000 day, tank volume (V) = 79,559 m^3.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{79559}{0.9000}$$

5. Step 5 — Evaluate:

   Q = 88399\ \text{m^3/day}

6. Step 6 — Check: returning Q = 88,399 m^3/day to

   $$V = Q \times \theta$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 176,798 — kept a factor of two that cancels in the correct rearrangement.
• 44,199 — dropped that same factor in the other direction.
• 97,239 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume

detention tank volume calculated from plant flow rate Given flow rate (Q) = 3,320 m^3/day; tank volume (V) = 377,106 m^3, determine the hydraulic detention time (theta) in day.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• Everything except theta is given, so isolate theta symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for theta:

   $$\theta = \dfrac{V}{Q}$$

3. Step 3

   $$List the givens: flow rate (Q) = 3,320 m^3/day, tank volume (V) = 377,106 m^3$$

4. Step 4 — Substitute the given values:

   $$\theta = \dfrac{377106}{3320}$$

5. Step 5 — Evaluate:

   $$\theta = 113.6\ \text{day}$$

6. Step 6 — Check: returning theta = 113.6 day to

   $$V = Q \times \theta$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 227.2 — kept a factor of two that cancels in the correct rearrangement.
• 56.7931 — dropped that same factor in the other direction.
• 124.9 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume

tank volume required for a target detention time in a storage tank Given flow rate (Q) = 6,100 m^3/day; hydraulic detention time (theta) = 7.1500 day, determine the tank volume (V) in m^3.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• Everything except V is given, so isolate V symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for V:

   $$V = Q\,\theta$$

3. Step 3 — List the givens: flow rate (Q) = 6,100 m^3/day, hydraulic detention time (theta) = 7.1500 day.

4. Step 4 — Substitute the given values:

   $$V = 6100\,7.1500$$

5. Step 5 — Evaluate:

   V = 43615\ \text{m^3}

6. Step 6 — Check: returning V = 43,615 m^3 to

   $$V = Q \times \theta$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 87,230 — kept a factor of two that cancels in the correct rearrangement.
• 21,808 — dropped that same factor in the other direction.
• 47,977 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume